I need help with #10 please
1 answer:
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The question is incomplete. The complete question is :
A sine function that has an amplitude of 16 units, a period of 5 units,a vertical displacement of 3 units up and a phase shift of 2.5 units left. Graph this function and show each step.
Solution :
The standard form of the sine function is
y = a sin (bx - c) + d
Here given :
Amplitude, a = 16
Period :
⇒
Phase shift :
∴
The vertical displacement : d = 3 units up
Now substituting a, b, c and d values in the standard form gives :
The graph is attached below.
Answer:
The value of MX is 5.5 units.
Step-by-step explanation:
As the complete question is not given, the complete question is found online and is attached herewith
The diagram is attached with the solution.
As the line ZM is dividing the angle in two parts thus as the two angles are equal thus
The sides are given as
XZ=11
YM=3.5
YZ=7
Now by the theorem of the angles, the bisected angles are such that sin(<YZM)=sin(<XZM)
so
So the value of MX is 5.5 units.
The two intersection points are (-2.79, -0.58) and (0.79, 6.58).
<h3>How to find the points of intersection?</h3>
Here we want to solve the system of equations:
y = 2x + 5
x² + y² = 36
To solve this, we need to replace the first equation into the second one:
x² + (2x + 5)² = 36
Now we can solve this for x:
x² + 4x² + 10x + 25 = 36
5x² + 10x - 11 = 0
This is a quadratic equation, to solve it we use the general formula:
So we have two solutions for x:
x = (-10 - 17.9)/10 = -2.79
x = (-10 + 17.9)/10 = 0.79
To get the y-values of the solutions, we evaluate the linear equation in these values of x:
y = 2*(-2.79) + 5 = -0.58
y = 2*( 0.79) + 5 = 6.58
Then the two intersection points are (-2.79, -0.58) and (0.79, 6.58).
If you want to learn more about intersection points:
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